2.5.3 Example C The equation Yk+1 – kyk = -e-1, ak = k and br = e-1, (2.97) has the particular solution 1 Yk = e-1 k(k +1).. (k +n)' (2.98) 82 n=0 58 Difference Equations Equation (2.97) is a special...

solve the Q(c) or explain the determain and the Q is complate .

Extracted text: 2.5.3 Example C The equation Yk+1 – kyk = -e-1, ak = k and br = e-1, (2.97) has the particular solution 1 Yk = e-1 k(k +1).. (k +n)' (2.98) 82 n=0 58 Difference Equations Equation (2.97) is a special case of Yk+1 – kyk = -e-Ppk, (2.99) where p is a constant. The general solution of this last equation is pn Yk = AT(k) +e¯Ppk (2.100) k(k +1). . · (k + n)’ n=0 where A is an arbitrary constant. Equation (2.97) corresponds to the situation where ρ= 1. Define P(k; p) and Q(k; p) as follows: P(k; p) = e-Ppk (2.101) k(k + 1) · . · (k +n)' n=0 Q(k; p) = r(k) – P(k; p). (2.102)